Question

${\cot }^{-1}\left\{\sqrt{\cos }\alpha \right\}-{\tan }^{-1}\left\{\sqrt{\cos }\alpha \right\}=x$, then $\sin x$ is equal to

• A. ${\tan }^2\left(\frac{\alpha }{2}\right)$
• B. ${\cot }^2\left(\frac{\alpha }{2}\right)$
• C. $\tan \alpha$
• D. $\cot \left(\frac{\alpha }{2}\right)$

Answer 1

The correct option is A ${\tan }^2\left(\frac{\alpha }{2}\right)$

Given, ${\cot }^{-1}\left\{\sqrt{\cos \alpha }\right\}-{\tan }^{-1}\left\{\sqrt{\cos \alpha }\right\}=x$

$\Rightarrow {\tan }^{-1}\left\{\frac{1}{\sqrt{\cos \alpha }}\right\}-{\tan }^{-1}\left\{\sqrt{\cos \alpha }\right\}=x$

$\Rightarrow {\tan }^{-1}\left\{\frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{\sqrt{\cos \alpha }}{\sqrt{\cos \alpha }}}\right\}=x$

$\Rightarrow \tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}$

Here $P=1-\cos \alpha$ and $B=2\sqrt{\cos \alpha }$

$H=\sqrt{(1-\cos \alpha {)}^2+4\cos \alpha }=\sqrt{(1+\cos \alpha {)}^2}=1+\cos \alpha$

Therefore, $\sin x=\frac{P}{H}=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{\sin }^2\frac{\alpha }{2}}{2{\cos }^2\frac{\alpha }{2}}={\tan }^2\frac{\alpha }{2}$